See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Ozonolysis of alkynes. When an alkyne is treated with ozone (O3) followed by reductive workup with Zn/H2O, the triple bond is cleaved oxidatively to give carbonyl compounds. For internal alkynes, ozonolysis cleaves the C≡C bond and each carbon of the triple bond becomes a carbonyl carbon. Step 1: Identify the substrate. CH3C≡C-CH3 is but-2-yne (2-butyne), an internal alkyne. Step 2: Apply ozonolysis mechanism. Ozone (O3) reacts with the triple bond in a [3+2] cycloaddition to form an ozonide intermediate. Upon reductive workup with Zn/H2O, the intermediate is cleaved. Step 3: Determine the products. For but-2-yne (CH3-C≡C-CH3), ozonolysis cleaves the C≡C bond. Each carbon of the triple bond, which originally bore a methyl group, becomes a carboxylic acid or, under reductive conditions (Zn/H2O), an aldehyde or ketone. Since each carbon of the triple bond in but-2-yne is bonded to one methyl group, each gives an acyl fragment CH3-C(=O)-, and these combine to give CH3-C(=O)-C(=O)-CH3 (butane-2,3-dione / diacetyl). This matches the product shown. Step 4: Eliminate other options. - (a) HNO3: Not a reagent used for this type of cleavage of alkynes to give 1,2-diketones. - (b) O2: Molecular oxygen does not perform ozonolysis-type cleavage of alkynes under these conditions. - (d) KMnO4: While KMnO4 can oxidize alkynes, it typically gives carboxylic acids under acidic/hot conditions, not the 1,2-diketone product via Zn/H2O workup as shown. Only ozone (O3) followed by Zn/H2O (reductive ozonolysis) cleaves the alkyne C≡C bond to give the observed 1,2-diketone product CH3CO-COCH3. Therefore, the correct answer is C.